Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m = 30;
int x[10005];
int y[10005];
vector<pair<int, int> > ansX;
vector<pair<int, int> > ansY;
void gauss(int v[], vector<pair<int, int> > &ans) {
  int cur = 0;
  for (int i = 0; i < m; ++i) {
    int idx = -1;
    for (int j = cur; j < n; ++j)
      if (v[j] >> i & 1) {
        idx = j;
        break;
      }
    if (idx == -1) continue;
    if (cur != idx) {
      v[cur] ^= v[idx], ans.push_back(make_pair(cur, idx));
      v[idx] ^= v[cur], ans.push_back(make_pair(idx, cur));
      v[cur] ^= v[idx], ans.push_back(make_pair(cur, idx));
    }
    for (int k = 0; k < n; ++k)
      if (k != cur && (v[k] >> i & 1)) {
        v[k] ^= v[cur], ans.push_back(make_pair(k, cur));
      }
    cur++;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", x + i);
  for (int i = 0; i < n; ++i) scanf("%d", y + i);
  gauss(x, ansX), gauss(y, ansY);
  for (int i = 0; i < m; ++i) {
    int bit = -1;
    for (int j = 0; j < m; ++j)
      if (x[i] >> j & 1) {
        bit = j;
        break;
      }
    if (bit == -1) continue;
    for (int j = 0; j < m; ++j)
      if ((x[j] >> bit & 1) ^ (y[j] >> bit & 1) & 1) {
        x[j] ^= x[i], ansX.push_back(make_pair(j, i));
      }
  }
  for (int i = 0; i < n; ++i)
    if (x[i] != y[i]) {
      printf("-1");
      return 0;
    }
  printf("%d\n", ansX.size() + ansY.size());
  for (int i = 0; i < ansX.size(); ++i)
    printf("%d %d\n", ansX[i].first + 1, ansX[i].second + 1);
  for (int i = (int)ansY.size() - 1; i >= 0; --i)
    printf("%d %d\n", ansY[i].first + 1, ansY[i].second + 1);
}